Some optimizations step by step:

The red condition is unnecessary, since the input has already been verified. When it comes to the orange test, it can be moved to the inner loop's condition:Code:for (a=2; a<=x; a++) { for (b=2; b<=x; b++) { if ((x>2) && ((x%2)==0) && (a<=b) && (a+b==x)) { cout << a << " and " << b << endl; } } }

Note that I reordered the loops, now the 'b' is outer.Code:for (b=2; b<=x; b++) { for (a=2; a<=b; a++) { if (a+b==x) { cout << a << " and " << b << endl; } } }

Going this way you can do what iMalc said eliminating the last condition.

Given 'a' in set: <2..b> there can be only one (or zero) 'a', such that (a + b = x), and this is (a = x - b):

If (b == x) then (a == 0), if (b == x - 1) then (a == 1). Both 0 and 1 are not primes:Code:for (b=2; b<=x; b++) { cout << x - b << " and " << b << endl; }

The code is much cleaner now, I think you will manage to implement IsPrime function. Just in case you get stuck:Code:for (b=2; b < x - 1; b++) { cout << x - b << " and " << b << endl; }

Code:for (b=2; b < x - 1; b++) { if (IsPrime(b) && IsPrime(x - b)) { cout << x - b << " and " << b << endl; } }